a 0.4 kg block rests on a desk. the coefficient of static friction is 0.2. You push the side of the block but do not have a spring scale to measure the force you use. the block does not move. which statement is true about the force of static friction?

Answers

Answer 1
Answer:

Answer:

D. It is no smaller than 0.78N

Explanation:

The question is incomplete. These are all the options :

A. It is no larger than 0.78N

B. It is 0.78N

C. It is 0.08N

D. It is no smaller than 0.78N

To solve this problem, we have the data of :

m=0.4kg

Where ''m'' is the mass of the block

μ = 0.2

Where ''μ'' is the coefficient of static friction

If we want to find the magnitude of the force of static friction we need to use the following equation :

F_(sf)= μ.F_(N) (I)

Where ''F_(N)'' is the normal force that the desk exerts on the block. Its magnitude is equal to the weight (because we suppose that the block rests horizontally on the desk).

The weight ''W'' can be calculated as :

W=m.g

Where ''m'' is the mass and ''g'' is the acceleration due to gravity.

The value of ''g'' is

g=9.81(m)/(s^(2))

The weight of the block is

W=(0.4kg).(9.81(m)/(s^(2)))

W=3.924N

Now, the weight is equal to the normal force ⇒

W=F_(N)=3.924N

Using the equation (I) :

F_(sf)=(0.2).(3.924N)

F_(sf)=0.7848N

The correct option is D. It is no smaller than 0.78N

Answer 2
Answer: Static friction is greater than Applied force

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Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it hits the desk.

Answers

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

 Here, we have been given the height through which the ball drops;

 Height of drop = 14.5m - 1.9m  = 12.6m

The right motion equation is;

      V²  = U² + 2gh

V is the final velocity

U is the initial velocity  = 0

g is the acceleration due to gravity  = 9.8m/s²

h is the height

  Now insert the parameters and solve;

       V² = 0² + 2 x 9.8 x 12.6

      V²  = 246.96

       V = √246.96  = 15.7m/s

Distance measurements based on the speed of light; used for objects in space

Answers

Answer : Yes, distance measurements based on the speed of light used for objects in space.

Explanation :  A light year is measurement  of distance  that light travel in a one year.

In a one year light travels 9460000000000 kilometer.

We know that, speed of light is 3*10^(8)\ m/s

and  time is 31536000 seconds in 1 year

so, distance= speed of light X time

Now, the light year is 9.4608*10^(15) meter

Example : The nearest star to earth is about 4.3 light year away.

A stone is dropped from a height of 7.70 m how fast is it going when it is 5.25 m above the ground

Answers

The velocity of the stone when it is 5.25 m above the ground is determined as 6.93 m/s.

Velocity of the stone

The velocity of the stone at the given displacement is calculated as follows;

K.E = ΔP.E

¹/₂mv² = mgh

v = √2gΔh

v = √[2(9.8)(7.7 - 5.25)]

v = 6.93 m/s

Thus, the velocity of the stone when it is 5.25 m above the ground is determined as 6.93 m/s.

Learn more about velocity here: brainly.com/question/4931057

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Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012W) pulses of light that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 5.0μm in diameter, with the pulse lasting for 4.0 ns with an average power of 2.0×1012W. We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity (in W/m2) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

Answers

Answer:

a. 80 J/cell b. 1.02 × 10²¹ W/m² c. 8.77 × 10¹¹ V/m d. 2.92 × 10³ T

Explanation:

a. We know that energy E = Pt where P = power and t = time

The total energy delivered to all the 100 cells is E = average power × time

average power = 2 × 10¹² W and time = 4 ns = 4 × 10⁻⁹ s

E = 2 × 10¹² W × 4 × 10⁻⁹ s = 8 × 10³ J

The energy per cell E₁ = E/100 = 8 × 10³ J/100 = 80 J/cell

b. Intensity, I = P/A where P = power per cell = 2 × 10¹² W/100 = 2 × 10¹⁰ W and A = area = πr². Since the cell is modeled as a disk of diameter d = 5.0μm, r = d/2 = 5.0 μm/2 = 2.5 μm = 2.5 × 10⁻⁶ m

I = P/A = P/πr² = 2 × 10¹⁰ W/π(2.5 × 10⁻⁶ m)² = 1.02 × 10²¹ W/m²

c. The intensity I = E²/2cμ₀ where E = maximum value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = 4π × 10⁻⁷ H/m

E = √(I2cμ₀) = √(2 × 1.02 × 10²¹ W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m) = 8.77 × 10¹¹ V/m

The maximum magnetic field B is gotten from E/B = c

B = E/c = 8.77 × 10¹¹ V/m/3 × 10⁸ m/s  = 2.92 × 10³ T

Ruff, the 50 cm tall Labrador Retriever stands 3 m from a plane mirror and looks at his image. What is Ruff’s image position and height?

Answers

Since the mirror is plane, the image will be formed behind the mirror. The distance will be the same as that of the distance of the object from the mirror and the height will just be the same.
So, Ruff's image will be 3 m behind the mirror and 50 cm tall.

The psychologist who worked with Peter and helped him associate rabbits with positive things was: a. Mary Covert-Jones
b. Ivan Pavlov
c. Elizabeth Loftus
d. John Watson

Knowing the formula for water and table salt is an example of:_________________memory.
a. episodic
b. procedural
c. semantic
d. echoic

The part of the brain that appears to be involved in the processing of short-term memories into long term is the:
a. hypothalamus
b. hippocampus
c. corpus callosum
d. cerebrum

Answers

1. Mary Cover-Jones 

2.Knowing the formula for water and table salt is an example of episodic memory.    so d

3.Hippocampus


Answer:

1) Mary Cover-Jones

2) episodic

3) hippocampus

Hope this helped man!